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Pp-Chain and CNO Cycle 10 – pp-chain and CNO cycle introduc)on to Astrophysics, C. Bertulani, Texas A&M-Commerce 1 pp chain reaction in the Sun introduc)on to Astrophysics, C. Bertulani, Texas A&M-Commerce 2 10.1 - Hydrogen burning – ppI chain As a star forms from a giant cloud in the interstellar medium (protostar), density and temperature increase in its center. Fusion of hydrogen (1H) is the first long term nuclear energy source that can ignite. With only hydrogen available (for example, a first generation star) the ppI chain (see previous slide) is the only possible sequence of reactions. (other reaction chains require catalyst nuclei). The ppI chain proceeds as follows: 2 Step 1: p + p → He This does not work because 2He is unstable + (10.1) p + p → d + e + ve 3 (10.2) Step 2: d + p → He d d 4 He This does not work because the d abundance is + → too low. d + p leads to rapid destruction of d. 3 4 Step 3: He + p → Li Does not work because 4Li is unstable 3 He + d → 4 He + n Does not work because d abundance is too low 3 He + 3 He → 4 He + 2p (10.3) Last reaction is OK because Y3He gets large as there is no other rapid destruction. introduc)on to Astrophysics, C. Bertulani, Texas A&M-Commerce 3 10.1.1 - S-factors for the pp chain All reactions in the pp-chain are charge-dependent. The cross section has the form given by Eq. (9.49) which can be cast in the form b − where 1 E σ ∝ e (10.4) A1A2 E b 31.28 Z Z A1/2 keV A = = 1 2 (10.5) A1 + A2 (10.6) Typical units for S(E) are keV barn. As we have written in Eq. (9.55), the reaction rate samples very low energy cross sections, or very low S-factors, through the Gamow Window E0 energy being very small. Then the integrand below around the Gamow window can be approximated by a Gaussian with energy width ΔE: E # b E & − -% + ( 8 −3/2 kT 8 −3/2 $ E kT ' < σ v >= (kT) ∫ σ (E)Ee dE = (kT) ∫ S(E)e dE πµ πµ where (10.7) 2 " b E % 1" E−E % 4 1/6 −$ + ' − $ 0 ' 2 2 5/6 kT 2$ ΔE/2 ' ΔE = E0kT = 0.23682 (Z1 Z2 A) T9 MeV e # E & ~ e # & 3 (10.8) (10.9) which leads to (9.55) after integration, as proven in the next two slides. introduc)on to Astrophysics, C. Bertulani, Texas A&M-Commerce 4 S-factors at very low energy The standard method for estimating such an integral is to find the energy that maximizes the exponential and expand around this peak in the integrand. This corresponds to solving ! $ 3/2 d # b E & 2E # + & = 0 ⇒ b = 0 (10.10) # & dE " E kT % kT Now carrying out a Taylor expansion around E0: ! $ 2 # b E & ! df $ 1 2 ! d f $ f (E) # & ~ f E E E E E = + 0 + − 0 # & + − 0 # 2 & +! (10.11) # & ( ) ( ) ( ) " E kT % "dE %E 2 "dE % 0 E0 ! $ df 1 2 # & = 0 ⇒ f (E) ~ f E + E − E f ''(E ) (10.12) "dE % ( 0 ) ( 0 ) 0 E0 2 1 2 - E-E f ''(E ) Thus 8 -f E ( 0) 0 < σ v >= (kT)−3/2 S(E)e ( 0)e 2 dE πµ ∫ 1 2 - E-E f ''(E ) 8 -f E ∞ ( 0) 0 ~ (kT)−3/2S(E )e ( 0) e 2 dE (10.13) πµ 0 ∫ 0 introduc)on to Astrophysics, C. Bertulani, Texas A&M-Commerce 5 S-factors at very low energy -f E ( 0) 4 −3/2 e < σ v >= (kT) S(E0 ) (10.14) µ f '' E ( 0 ) 3b 3 But E0 b 3E0 f ''(E ) f (E ) and 0 = 5/2 = 0 = + = 4E 2E kT kT E kT 0 0 0 (10.15) (10.16) Thus, we get Eq. (9.55) as promised: ! $2 16 1 1 3E0 < σ v >= S(E0 )# & 9 3 µ 2παZ1Z2 " kT % introduc)on to Astrophysics, C. Bertulani, Texas A&M-Commerce 6 S-factors at very low energy Better (and this is often done for experimental data) one expands S(E) around E = 0 as powers of E to second order: 1 S(E) = S(0) + ES'(0) + E2S''(0) (10.17) 2 If one integrates this over the Gamov window, one finds that one can use equation of previous slide when replacing S(E0) with the effective S-factor Seff ' 5 S'(0) ! 35 $ 1 S''(0) ! 2 89 $* Seff = S(0))1+ + #E0 + kT&+ #E0 + E0kT&, ( 12τ S(0) " 36 % 2 S(0) " 36 %+ 3E0 (10.18) with τ = and E0 as location of the Gamov window. kT introduc)on to Astrophysics, C. Bertulani, Texas A&M-Commerce 7 The ppI chain Let us start by looking at the nuclear chart Z 3He 4He 2 In our notation, we call this as a 1H 2H “chain” of reactions, or 1 reaction chain: 0 1 N “bottle neck” (p,e+) (p,γ) (3He,2p) 1H d 3He 4He The first reaction in the chain is p + p à deuteron (d) + e+ + ν. The cross section for it is very small. The other reactions have to wait until this reaction occurs. That is why this reaction is called a “bottleneck”. introduc)on to Astrophysics, C. Bertulani, Texas A&M-Commerce 8 The p + p à d reaction The p + p à deuteron (d) + e+ + ν cross section is very small because the protons have to overcome their mutual repulsion, tunnel to a distance of about 1 fm. Then one of them has to decay to neutron + neutrino and the neutron has to be captured by the proton to form the deuteron. The cross-section for this reaction for protons of energy below 1 MeV is very small, of the order of 10−23 b. In fact, the effective energy for this reaction in the Sun is about 20 keV and this value is much lower. Another possible reaction forming the deuteron is p + e- + p à d + n, but the frequency with which it occurs is about 400 times smaller than the pp reaction. -25 The S-factor for the pp reaction is Spp(0) = 4 x 10 MeV b and the reaction rate for it is found to be 9 # & 1 2 1.15×10 2 2 3.38 3 1 r = n σ v = X ρ exp%− ( cm− s− (10.19) pp 2 p pp T2/3 % T1/3 ( 9 $ 9 ' where X is the hydrogen mass fraction. The higher order correction leads to -1 S’pp(0) = (dSpp/dE)E=0 = 11.2 × Spp(0) MeV . See Reviews of Modern Physics 85, 195 (2011). introduc)on to Astrophysics, C. Bertulani, Texas A&M-Commerce 9 The p + p à d reaction The rate of change for proton number because of the pp reaction is given by the dn p n p = − (10.20) dt τ p where τp is the proton half-life due to the reaction. The lifetime is then n p n p (10.21) τ p = − = dn p / dt 2rpp where the factor two is because two protons are destroyed in the process. -3 Using T9 = 0.15, X = 0.5 and ρ = 100 g cm in Eq. (10.19) and using Eq. (10.21) one 9 gets τp ~ 5 x 10 years. This is a reasonable estimate for the lifetime of the sun. Note: it is customary to call Spp by S11, meaning A = 1 for each of the two protons. For example, the S-factor for the pd reaction is called S12. introduc)on to Astrophysics, C. Bertulani, Texas A&M-Commerce 10 p + d à 3He + γ The S-factor for this reaction is S12(0) = 0.241 eV b. In the figure we show the world data (as of 2011) for the S- factor for this reaction. The curves are theoretical calculations and/or experimental fits. The band is associated to the experimental error bars. introduc)on to Astrophysics, C. Bertulani, Texas A&M-Commerce 11 3 3 4 He + He à He + 2p The S-factor for this reaction needs to include higher derivatives of S for a better precision of the reaction rate calculation, as in Eq. (10.17). In the figure we see that the laboratory screening causes the S- factor to deviate from the desired “bare” S- factor. It is thus necessary to include screening corrections. The table below summarizes the fits for this reaction Rev. Mod. Phys. 85, 195 (2011). introduc)on to Astrophysics, C. Bertulani, Texas A&M-Commerce 12 10.2 - Reaction networks with a bottle-neck Consider a schematic chain of proton captures shown below: (p, γ) (p, γ) (p, γ) (p, γ) 1 2 3 4 Slow bottle neck We will assume that: • Y1 ~ const because the depletion is very slow due to the “bottle neck” • Capture rates are constant (Yp ~ const because of the large “reservoir”, abundant protons in the environment) The abundance of nucleus 2 evolves according to the equations dY 1 2 (10.22) λ = Y ρ N < σ v > (10.23) = Y1λ12 −Y2λ23 12 1+δ p A 1→2 dt p1 The first term on the rhs of Eq. (10.22) is due to production of 2, the second term due to destruction of 2. introduc)on to Astrophysics, C. Bertulani, Texas A&M-Commerce 13 Reaction networks with a bottle-neck We have assumed that Y1 ~ const and Yp ~ const.
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